To solve the problem, we will first calculate the buoyant force acting on the block when it is immersed in water and then in another liquid. We will use this information to find the relative density of the block and the liquid.
### Step-by-Step Solution
1. **Identify the Given Data:**
- Weight of the block in air, \( W_{\text{air}} = 15 \, \text{N} \)
- Weight of the block in water, \( W_{\text{water}} = 12 \, \text{N} \)
- Weight of the block in another liquid, \( W_{\text{liquid}} = 13 \, \text{N} \)
2. **Calculate the Buoyant Force in Water:**
The buoyant force \( F_b \) in water can be calculated as:
\[
F_b = W_{\text{air}} - W_{\text{water}} = 15 \, \text{N} - 12 \, \text{N} = 3 \, \text{N}
\]
3. **Calculate the Buoyant Force in the Other Liquid:**
The buoyant force \( F_b \) in the other liquid can be calculated as:
\[
F_b = W_{\text{air}} - W_{\text{liquid}} = 15 \, \text{N} - 13 \, \text{N} = 2 \, \text{N}
\]
4. **Determine the Volume of the Block:**
The buoyant force is also given by the formula:
\[
F_b = V \cdot \rho_{\text{fluid}} \cdot g
\]
where \( V \) is the volume of the block, \( \rho_{\text{fluid}} \) is the density of the fluid, and \( g \) is the acceleration due to gravity.
For water:
\[
3 \, \text{N} = V \cdot \rho_{\text{water}} \cdot g
\]
Since the density of water \( \rho_{\text{water}} \) is \( 1000 \, \text{kg/m}^3 \) and \( g \approx 9.81 \, \text{m/s}^2 \):
\[
3 = V \cdot 1000 \cdot 9.81
\]
Solving for \( V \):
\[
V = \frac{3}{1000 \cdot 9.81} \approx 0.000305 \, \text{m}^3
\]
5. **Calculate the Density of the Block:**
The density of the block \( \rho_b \) can be calculated using its weight in air:
\[
W_{\text{air}} = V \cdot \rho_b \cdot g
\]
Rearranging gives:
\[
\rho_b = \frac{W_{\text{air}}}{V \cdot g} = \frac{15}{0.000305 \cdot 9.81} \approx 5000 \, \text{kg/m}^3
\]
6. **Calculate the Relative Density of the Block:**
The relative density \( RD_b \) of the block is given by:
\[
RD_b = \frac{\rho_b}{\rho_{\text{water}}} = \frac{5000}{1000} = 5
\]
7. **Calculate the Density of the Liquid:**
Using the buoyant force in the other liquid:
\[
2 \, \text{N} = V \cdot \rho_{\text{liquid}} \cdot g
\]
Rearranging gives:
\[
\rho_{\text{liquid}} = \frac{2}{V \cdot g} = \frac{2}{0.000305 \cdot 9.81} \approx 600 \, \text{kg/m}^3
\]
8. **Calculate the Relative Density of the Liquid:**
The relative density \( RD_l \) of the liquid is given by:
\[
RD_l = \frac{\rho_{\text{liquid}}}{\rho_{\text{water}}} = \frac{600}{1000} = 0.6
\]
### Final Answers:
- (i) The relative density of the block is \( 5 \).
- (ii) The relative density of the liquid is \( 0.6 \).
